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STRUCTURE OF COBALT ISOTOPES
By Prof.Lefteris Kaliambos (Natural Philosopher in New Energy). (August 2014) Unfortunately the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories which could not lead to the nuclear structure. Under this physics crisis in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003) which led to my discovery of the new structure of protons and neutrons given by New structure of proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons........ New structure of neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons........ Here one sees that the 9 charged quarks in proton and the 12 ones in neutron give the charge distributions in nucleons for revealing the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism (See my papers of nuclear structure in FUNDAMENTAL PHYSICS CONCEPTS . Naturally occurring cobalt (Co) is composed of 1 stable isotope, Co-59. 28 radioisotopes have been characterized with the most stable being Co-60 with a half-life of 5.2714 years, Co-57 with a half-life of 271.8 days, Co-56 with a half-life of 77.27 days, and Co-58 with a half-life of 70.86 days. All of the remaining radioactive isotopes have half-lives that are less than 18 hours and the majority of these have half-lives that are less than 1 second. This element also has 11 meta states, all of which have half-lives less than 15 minutes. The isotopes of cobalt range in atomic weight from Co-47 to Co-75. The primary decay mode for isotopes with atomic mass unit values less than that of the most abundant stable isotope, Co-59, is electron capture and the primary mode of decay for those of greater than 59 atomic mass units is beta decay. The primary decay products before Co-59 are iron isotopes and the primary products after are nickel isotopes. Radioactive isotopes can be produced by various nuclear reactions. For example, the isotope Co-57 is produced by cyclotron irradiation of iron. The principal reaction involved is the (d,n) reaction Fe-56 + 2H → n + Co-57. ' ' WHY Co-59 WITH S = -7/2 IS A STABLE NUCLIDE After a careful analysis of the structure of atomic nuclei I discovered that the beta decay is due to the fact that in unstable nuclei there exist single horizontal pn bonds of weak binding energy leading to the beta decay. For example in my paper STRUCTURE AND BINDING OF H3 AND He3 using the diagram of the structure of the H3 one sees that it is unstable because the two neutrons make single np bonds, while the He3 is stable because the one neutron between the two protons makes two np bonds per neutron. On the other hand the pp repulsions of long range lead to the instability when we have a small number of pn bonds per nucleon. For comparing the structure of the stable Co-59 with the unstable structure of Co-56 you may read my STRUCTURE OF Co-59 AND Co-56 . A careful analysis of such a comparison shows that the stable structure of Co-59 is due to the enough number of extra neutrons. It has 5 extra neutrons which make two bonds per neutron able to increase enough the energies of bonds for overcoming the pp and nn repulsions. Whereas the unstable Co-56 with S = +4 has only two extra neutrons unable to give enough energies to pn bonds for overcoming the pp and nn repulsions. ' ' NUCLEAR STRUCTURE OF Co-60 WITH S = +5 In the presence of 2 more extra neutrons of positive spins than those of Co-56, and of 2 more extra neutrons of opposite spins giving S = 0 we get the unstable structure of Co-60 with S = +5. That is S = +4 + 2(+1/2) + 0 = +5 ' ' NUCLEAR STRUCTURE OF Co-61, Co-63, Co-65, Co-67, Co-69, Co-71, Co-73 AND Co-75 WITH S =-7/2 ' After a careful analysis I found that the structure of the above unstable nuclides is based on the structure of Co-59 with S =-7/2. For example the Co-75 with S=-7/2 has 16 more extra neutrons than those of Co-59 with opposite spins which make single weak horizontal bonds leading to the decay. ' ' '''NUCLEAR STRUCTURE OF Co-57 AND Co-55 WITH S = -7/2 ' In the absence of neutrons with opposite spins we see that the structure of the above unstable nuclides is based also on the structure of Co-59 with S = -7/2. For example in the absence of n29(+1/2) and n32(-1/2) we get the structure of Co-57 with the same S =-7/2. Similarly in the absence of the next n28(+1/2 )and n31(-1/2) we get the structure of Co-55 with the same S =-7/2. ' ' 'NUCLEAR STRUCTURE OF Co-53, Co-51, Co-49 AND Co-47 WITH S =-7/2 ' Here we see that the number of neutrons is smaller than the number of 27 which gives the structure of Co-54 . However in the absence of n29, n32, n28 and n31 we have in this structure the one extra neutron of negative spin. In other words the total spin of a new structure of Co-54 having 27 protons and 27 neutrons is given by S = -7/2 +1/2 = -3 Therefore in the absence of one more neutron of positive spin we get the structure of Co-53 with S = -7/2 given by S = -3 - 1(+1/2) = -7/2 Under this condition the structure of Co-51, Co-49 and Co-47 is based on the structure of Co-53 with S=-7/2. For example in the structure of Co-47 with S = -7/2 we have 6 absent neutrons of opposite spins more than those of Co-53 with the same S =-7/2. ' ' '''NUCLEAR STRUCTURE OF Co-68 WITH S = -7, THE Co-70 WITH S = -6 AND THE Co-72 WITH S = -6 Here we see that the unstable nuclides have negative spins like the new structure of Co-54 with S = -3 which differs from the total spin of the structure of the well-known Co-54 with S = 0. Under this condition the structure of the above nuclides is based on this new structure of Co-54 with S = -3. For example the Co-72 with S =-6 has 6 more extra neutrons of negative spins and 12 more extra neutrons of opposite spins giving S = 0. That is S = -3 + 6(-1/2) + 0 = -6 . ' ' STRUCTURE OF Co-54 WITH S = 0 In the following diagram of Co-54 with S = 0 you see that it has the core of Mg-24 with 6 horizontal planes of opposite spins like the +HP1, -HP2, +HP3, -HP4, +HP5 and -HP6 . It consists of 26 deuterons with the total S= 0 existing from p1n1 to p26n26 which form the two blank positions at +HP1 and -HP6 for receiving the p27(+1/2) and the n27(-1/2). Since the 4 alpha particles of the +HP3 and -HP4 give S=0 we see that the deuterons of +HP1, -HP2 +HP5 and –HP6 give also S=0 because +HP1 has 3 deuterons with S = +3 -HP2 has 4 deuterons with S = -4 +HP5 has 4 deuterons with S = +4 -HP6 has 3 deuterons with S = -3 Then adding the zero spin of p27(+1/2) and n27(-1/2) we get the total spin S = 0 of the structure of Co-54. ' ' DIAGRAM OF THE UNSTABLE Co-54 WITH S=0 In this structure the deuterons n17p17 and p18n18 of the alpha particle are not shown because they are in front of p5n5 and n7p7 respectively. Also the p19n19 and n20p20 of the symmetrical alpha particle are not shown because they are behind the n6p6 and p8n8 respectively. ' ' ' n27…….p12.........n12..........p26' ' -HP6 n11........p11....... n26 ' ' p23.......n10..........p10.......... n24' ' +HP5 n23........p9...........n9.........p24 ' ' n14........p8............n8............p16' ' -HP4 p14........n7...........p7.........n16 ' ' p13........n6........... p6............n15' ' +HP3 n13.........p5..........n5.........p15 ' ' n21……...p4............n4………..p22' ' -HP2 p21……….n3...........p3 ……..n22 ' ' p25........n2............p2………..n26' ' +HP1 n25........p1...........n1.........p27 ' NUCLEAR STRUCTURE OF Co-58, Co-62, Co-64, Co-66 AND Co-74 ' After a careful analysis I found that the structure of the above unstable nuclides is based on the structure of Co-54 with S = 0. For example the unstable structure of Co-58 with S = +2 has 4 extra neutrons of positive spins giving the total S = +2. That is S = 0 + 4(+1/2) = +2 ' ''' '''NUCLEAR STRUCTURE OF Co-52, Co-50 AND Co-48 WITH S = +6 In the absence of neutrons and after a detailed analysis of the above unstable nuclides we found that the structure of them is based on a structure with 27 protons and 27 neutrons in which significant arrangements of nucleons give S = +6. In this case the n27(-1/2) of the diagram of Co-54 with S = 0 becomes n27(+1/2) because it goes from the -HP6 to the +HP1 to form the deuteron p27n27 with S = +1.That is, this change of spin gives S = +1. Also the p12n12 and n26p26 change their spins from S = -2 to S = +2 because they go from the -HP6 to the +HP5 to form bonds at symmetrical positions existing in front of p9n9 and behind the n10p10. Note that this change of spins gives S = +4. Moreover the movement of the deuteron n11p11 gives S=+1 because it changes the spin from S=-1 to S=0. Especially it goes at positions existing in front of p1 and n3 in order to make a vertical n11p11 with S = 0 In other words under this condition the total spin is given by S = +1 +4 +1 =+6 So in the absence of 2 neutrons with opposite spins we get the structure of Co-52 with S = +6. Similarly in the absence of 4 or 6 neutrons of opposite spins we get the structure of Co-50 and of Co-48 respectively. Category:Fundamental physics concepts